\section{答案}
\subsection{例题}
\begin{enumerate}
    \item 略
    \item \(\begin{aligned}
        -\frac{1}{(n+1)^{2}}
    \end{aligned}\)
    \item \(\begin{aligned}
        \int_0^{\pi/2}\frac{\cos x\sin x}{a^2\sin^2x+b^2\cos^2x}\mathrm{d}x=\begin{cases}\dfrac{\ln|a|-\ln|b|}{a^2-b^2},&a\neq b\\\dfrac{1}{2a^2},&a=b\end{cases}.
    \end{aligned}\)
    \item \(\begin{aligned}
        \frac{\ln\cot(\pi/8)}{\sqrt2}.
    \end{aligned}\)
\end{enumerate}
\subsection{计算题}
\begin{enumerate}[(1)]
    \item $\begin{aligned}&\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\cdots+\frac{n}{n^2+n^2}\right)\\&=\lim_{n\to\infty}\frac{1}{n}\left(\frac{1}{1+(1/n)^2}+\frac{1}{1+(2/n)^2}+\cdots+\frac{1}{1+(n/n)^2}\right)\\&=\int_0^1\frac{\mathrm{d}x}{1+x^2}=\frac{\pi}{4}.\end{aligned}$
    \vspace{1em}
    \item $\lim\limits_{n\to\infty}\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}=\int_0^1x^p\mathrm{d}x=\frac1{p+1}\text{。}$
    \vspace{1em}
    \item $\int_{1}^{4}\frac{x+1}{\sqrt{x}}\mathrm{d}x=\int_{1}^{4}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\mathrm{d}x=\frac{20}{3}。$
    \vspace{1em}
    \item $\begin{aligned}\int_{0}^{\sqrt{3}}x\arctan x\mathrm{d}x&=\frac{1}{2}\int_{0}^{\sqrt{3}}\arctan x\mathrm{d}x^{2}=\frac{\pi}{2}-\frac{1}{2}\int_{0}^{\sqrt{3}}\frac{x^{2}}{1+x^{2}}\mathrm{d}x=\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\end{aligned}$
    \vspace{1em}
    \item $\int_{-1}^{0}(2x+1)\sqrt{1-x-x^{2}}\mathrm{d}x=-\int_{-1}^{0}\sqrt{1-x-x^{2}}\mathrm{d}(1-x-x^{2})=0.$
    \vspace{1em}
    \item $\int_{0}^{a}\ln(x+\sqrt{a^{2}+x^{2}})\mathrm{d}x=(x\ln(x+\sqrt{a^{2}+x^{2}})-\sqrt{a^{2}+x^{2}})|_{0}^{a}=a\ln(1+\sqrt{2})a+a(1-\sqrt{2}).$
    \vspace{1em}
    \item $\int_0^{\ln2}\sqrt{\mathrm{e}^x-1}\mathrm{d}x\stackrel{\sqrt{\mathrm{e}^x-1}=u}{=}2\int_0^1\frac{u^2}{1+u^2}\mathrm{d}u=2-\frac{\pi}{2}\int_0^{\ln2}\sqrt{\mathrm{e}^x-1}\mathrm{d}x\stackrel{\sqrt{\mathrm{e}^x-1}=u}{=}2\int_0^1\frac{u^2}{1+u^2}\mathrm{d}u=2-\frac{\pi}{2}$
    \vspace{1em}
    \item $\begin{aligned}
\int_0^{\pi/4}\frac{\mathrm{d}x}{\cos x}& =\int_0^{\pi/4}\frac{\mathrm{d}\sin x}{1-\sin^2x}=\int_0^{\sqrt{2}/2}\frac{\mathrm{d}t}{1-t^2}  \\
&=\frac12(\int_0^{\sqrt{2}/2}\frac{\mathrm{d}t}{1+t}+\int_0^{\sqrt{2}/2}\frac{\mathrm{d}t}{1-t}) \\
&=\left.\frac12\ln\frac{1+t}{1-t}\right|_0^{\sqrt{2}/2}=\left.\frac12\ln\left(3+2\sqrt{2}\right).\right. 
\end{aligned}$
    

\end{enumerate}
\subsection{证明题}
\begin{enumerate}
    \item \begin{proof}
利用$t=x+h$进行换元，得到：
$$
\begin{aligned}
\int_a^bf(x+h)\mathrm{d}x& =\int_{a+h}^{b+h}f(t)\mathrm{d}t  \\
&=\int_{a}^{b}f(\begin{array}{c}t\\\end{array})\mathrm{d}t+\int_{b}^{b+h}f(t)\mathrm{d}t-\int_{a}^{a+h}f(t)\mathrm{d}t.
\end{aligned}
$$
由积分中值定理得到：
$$
\int_{b}^{b+h}f(t)\mathrm{d}t=f(\xi)h \quad (\xi \text{在} b \text{和} b+h \text{之间})
$$
$$
\int_{a}^{a+h}f(t)\mathrm{d}t=f(\eta)h \quad (\eta \text{在} a \text{和} a+h \text{之间})
$$
因此
$$
\frac1h\int_a^b(f(x+h)-f(x))\mathrm{d}x=f(\xi)-f(\eta).
$$
当$h\rightarrow 0$时，
$$
\lim_{h\to0}f(\xi)=f(b),\quad\lim_{h\to0}f(\eta)=f(a).
$$
原命题得证.
\end{proof}
\item \begin{proof}
    第一步先证明
    \[
    \int_{0}^{\pi}\frac{\sin\left(n+\frac{1}{2}\right)x}{\sin\frac{x}{2}}\mathrm{d}x=\pi(n\in \mathrm{N}^*)
    \]
    因为
    \[
\frac{\sin(n+1/2)x}{\sin(x/2)}=1+2(\cos x+\cos2x+\cdots+\cos nx).
    \]
    对两边从$0$到$2\pi$积分即可结果，
    \[\begin{aligned}
        \sin^2x-\sin^2y=\sin(x-y){\sin(x+y)},
    \end{aligned}\]
    可得
    \[
    \begin{aligned}\int_0^\pi\left(\frac{\sin((n+1)x/2)}{\sin(x/2)}\right)^2\mathrm{d}x&=\int_0^\pi\left(\frac{\sin(nx/2)}{\sin(x/2)}\right)^2\mathrm{d}x+\int_0^\pi\frac{\sin{\left((n-\frac{1}{2})x\right)}\sin(x/2)}{(\sin{x/2)^2}}\mathrm{d}x\\&=\int_0^\pi\left(\frac{\sin(nx/2)}{\sin(x/2)}\right)^2\mathrm{d}x+\pi\end{aligned}
    \]
    递推可得结果为\(n\pi\)
\end{proof}
\end{enumerate}

